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fgiga [73]
2 years ago
15

Area of a trapezium​

Mathematics
2 answers:
melamori03 [73]2 years ago
7 0

Answer:

The area of a trapezium is computed with the following formula: Area = 1 2 × Sum of parallel sides × Distance between them.

Step-by-step explanation:

Hope it help! :)

Vlad [161]2 years ago
5 0
You didn’t give any numbers so I’ll show you a picture of how to calculate it:)

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A restaurant has fixed costs of $147.50 per day and an average unit cost of $5.75 for each meal served. If a typical meal costs
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Answer:

The answer is 118.

Step-by-step explanation:

First we need to subtract the meal cost from average unit cost for each meal:

7-5.75=1.25

A restaurant profits 1.25$ for each meal. If we divide fixed cost to profit for each meal:

147.5/1.25=118

The restaurant have to sell 118 meals at least each day for the owner to break even

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What are the x and y intercepts of 2x-3y=6?
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X=67/2
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x and y intercepts
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The length of a rectangle is 3 times its width. If the length were increased by 2 inches and the width by 1 inch, the new perime
Dennis_Churaev [7]
Pretty sure it’s the second option because length would be 3w+2
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2 years ago
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Perform the indicated row operations, then write the new matrix.
Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]

Operations:

2R_1 + R_2 ->R_2

-3R_1 +R_3 ->R_3

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

Step-by-step explanation:

The first operation:

2R_1 + R_2 ->R_2

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by 2

2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}0&5&7|1\end{array}\right]

The second operation:

-3R_1 +R_3 ->R_3

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by -3

-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]

Hence, the new matrix is:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

3 0
3 years ago
Read 2 more answers
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